06A06-HeightOfAnElementInAPoset.tex

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\pmtitle{height of an element in a poset}
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Let $P$ be a poset.  Given any $a\in P$, the lower set $\down a$ of $a$ is a subposet of $P$.  Call the height of $\down a$ less 1 the \emph{height} of $a$.  Let's denote $h(a)$ the height of $a$, so $$h(a)=\operatorname{height}(\down a)-1.$$  From this definition, we see that $h(a)=0$ iff $a$ is minimal and $h(a)=1$ iff $a$ is an atom.  Also, $h$ partitions $P$ into equivalence classes, so that $a$ is equivalent to $b$ in $P$ iff $h(a)=h(b)$.  Two distinct elements in the same equivalence class are necessarily incomparable.  In other words, the equivalence classes are antichains.  Furthermore, given any two equivalence classes $[a],[b]$, set $[a]\le[b]$ iff $h(a)\le h(b)$, then the set of equivalence classes form a chain.

The height function of a poset $P$ looks remarkably like the rank function of a graded poset: $h$ is constant on the set of all minimal elements, and $h$ is isotone (preserves order).  When is $h$ a rank function (the additional condition being the preservation of the covering relation)?  The answer is given by a chain condition imposed on $P$, called the \emph{Jordan-Dedekind chain condition}:
\begin{quote}
(*) In a poset, the cardinalities of two maximal chains between common end points must be the same.
\end{quote}
Suppose for each $a\in P$, $h(a)$ is finite and $P$ has a unique minimal element $0$.  Then $P$ can be graded by $h$ iff (*) is satisfied.  More generally, if we drop the assumption of the uniqueness of a minimal element, then $P$ can be graded by $h$ iff any two maximal chains ending at the same end point have the same length.

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